Credits: Four. School. Level 2 Mathematics and Statistics. Practice Assessment B (2.7) Apply calculus methods in solving problems.
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1 Name: Teacher: NZAMT 2 Credits: Four School Level 2 Mathematics and Statistics Practice Assessment B (2.7) Apply calculus methods in solving problems 5 credits You should answer ALL parts of ALL questions in this booklet. You should show ALL your working. If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly number the question. Check that this booklet has pages 2 10 in the correct order and that none of these pages is blank. YOU MUST HAND THIS BOOKLET TO YOUR TEACHER AT THE END OF THE ALLOTTED TIME
2 2 You are advised to spend 60 minutes answering the questions in this booklet. QUESTION ONE: (a) Find the gradient of the function y = 12 3x x 3 when x= -1 (b) (i) By differentiating f(x) = 3x 4 6x 2 + 8, determine if the graph is increasing or decreasing when x = 2 (ii) Find the coordinates of the turning points on the graph of f(x) = 3x 4 6x (iii) Give the value of f ''(x) for the turning point with the largest x coordinate found in part(ii), and explain what this means.
3 3 (c) The graph of y = f (x) below is made up of a straight line and a parabola. y x -2-4 On the axes below, sketch the graph of the gradient function. y x -2-4
4 4 (d) A suitcase in the shape of a cuboid is twice as long as it is high. Certain airlines will allow a total of the length, width and height to be 180cm before making an extra charge. Find the maximum volume a case can have before there is an extra charge and give its dimensions.
5 5 QUESTION TWO (a) The gradient function is given by f (x) = 6x 2 + 2x 8. What is f(x) if (2,-8) lies on the curve? (b) Give an equation for the tangent at x = 2 for the function f(x) = x 5 6x (c) A remote control car starts moving in a straight line from a fixed point. The velocity in metres per second is given by the formula v = 3t 2 4t 8 (i) what is the initial acceleration? (ii) how far does the car travel in the 3rd second? (d) An object is moving in a straight line. It has an acceleration of a m sec -2 where a = ct + d and t is the time since it passed a fixed point on the line. What is the maximum number of times the object changes direction? Explain your answer.
6 6 (e) An object is launched from ground level. The velocity v in metres per second of an object for the first 8 seconds is given by the formula v = 40 10t where t is time measured in seconds. The graph of the velocity is shown below. The distance is the height above ground level. v t Give a full description of the motion of the object after 6 seconds. You could include in your description: the direction, position, velocity, and acceleration of the object providing values if appropriate. \
7 7 QUESTION THREE: (a) The area of a lake in km 2 is given as a function of time t (days) after a storm A = 150t 0.3t 4 (i) Differentiate A and find the rate at which the area is increasing after 4 days (ii) When does the lake reach its maximum area and give the maximum area? (b) For the curve y = x3 3 7x x + 1 (i) Use calculus methods to find the values of x where the curve is decreasing.
8 8 (ii) Find the value of x where the curve in part (i) y = x3 3 7x x + 1 is decreasing the fastest. (c) A rectangular sheet of metal having a perimeter of 100cm is rolled into a cylinder. What should the dimensions of the metal sheet be for the cylinder to have a maximum volume? (Formulae: Circumference = 2πr, Area circle = πr 2,Volume of a cylinder =πr 2 h)
9 9 Extra paper for continuing your answers, if required. Clearly number the question. Question number
10 10 Extra paper for continuing your answers, if required. Clearly number the question. Question number
11 ASSESSMENT SCHEDULE Mathematics and Statistics (2.7): Apply calculus methods in solving problems 11 Achievement Merit Excellence Apply calculus methods in solving problems involves: selecting and using methods demonstrating knowledge of calculus concepts and terms communicating using appropriate representations. Apply calculus methods using relational thinking, in solving problems must involve one or more of: selecting and carrying out a logical sequence of steps connecting different concepts and representations demonstrating understanding of concepts forming and using a model and relating findings to a context, or communicating thinking using appropriate mathematical statements. Apply calculus methods using extended abstract thinking, in solving problems must involve one or more of: devising a strategy to investigate a situation demonstrating understanding of abstract concepts developing a chain of logical reasoning, or proof forming a generalisation and using correct mathematical statements, or communicating mathematical insight.
12 Evidence Statement 12 One Expected Coverage Achievement(u) Merit(r) Excellence(t) NØ = No response; no relevant evidence. N1 = a valid attempt at ONE question. N2 = ONE question demonstrating limited knowledge of calculus techniques.(1u) A3 = TWO of u. A4 = THREE of u. M5 = ONE of r. M6 = TWO of r. E7 = ONE of t, with minor errors ignored. E8 = 2 of t (a) f (x)' = - 6x + 0.9x 2 grad = 6.9 Derivative found and finds the gradient at the point where x = -1. (b)(i) dy dx = 12x3 12x when x=2, gradient=72 the graph is increasing differentiates and substitutes in x=2 gradient found and states the curve is increasing (b)(ii) dy dx = 12x3 12x = 0 equates derivative to 0 and partially solves x = -1,0,1 so there is a stationary points at (-1,5),(0,8),(1,5) finds all three stationary points (b)(iii) y = 36x 2-12, at x=1, y =24 >0 means the point is minimum or that that the gradient is increasing gives second derivative allow consistency evaluates y for x=1 gives an explanation for positive y Graph of derivative 2 y derivative drawn, with the derived value of 2 when x<0, and an x- intercept at 1 (the graph does not need to be continuous at x=0) (c) 1 x finds an expression for the volume and derives equates derivative to 0 and solves gives the maximum volume (d) Volume is 0.192m 3.
13 = x + 2x + y y = 180 3x VolumeV = 2x 2 y V = 2x 2 (180 3x) = 360 x 2 6x 3 V ' = 720 x 18x 2 = 0 x = 40 so max V = cm 3 or 192 L Dimensions of suitcase are width 40cm, length 80cmandheight 60cm.
14 Two Expected Coverage u r t NØ = No response; no relevant evidence. N1 = a valid attempt at ONE question. N2 = ONE question demonstrating limited knowledge of calculus techniques.(1u) A3 = TWO of u. A4 = THREE of u. M5 = ONE of r. M6 = TWO of r. E7 = ONE of t, with minor errors ignored. E8 = 2 of t. 14 (a) y = 2x 3 + x 2 8x + c -8 = 2(2) 3 + (2) 2 8(2) + c c = -12 Anti-derivative found, c evaluated so y = 2x 3 + x 2 8x 12 (b) f '(x) = 5x 4 18x 2 point (2,-9), gradient = 8 derivative found and evaluated at x=2 equation of tangent found tangent equation y = 8x 25 (c)(i) a=6t-4 when t=0, a=-4 Expression for acceleration and gives the initial acceleration (c)(ii) s = t 3-2t 2-8t in the third second the car travels s(3)-s(2)=1metre anti-derives, and finds the distance for t=2 or t=3 finds the distance travelled in the 3 second (d) direction changes a maximum of two times. Change of direction implies v = 0. finds the number of time the direction changes explains the answer. v is represented by a quadratic function which has a maximum of 2 solutions. (e) when t=6, s = 40t 5t 2, s=60 m above the ground v=-20ms -1,,the velocity is 20ms -1 and the object is travelling towards the ground a=-10ms -2, the object is accelerating towards the ground anti-derives to find the distance function and evaluates it when t=6 derives to find the acceleration describes two of aspects in terms of the objects motion in relation to the ground, numercial quantities for at least one. describes s, v and a fully by giving values and a description in terms of the relationships 1t=fully correct for 2 aspects 2t = correct for all 3 aspects
15 15 Three Expected Coverage u r t NØ = No response; no relevant evidence. N1 = a valid attempt at ONE question. N2 = ONE question demonstrating limited knowledge of calculus techniques.(1u) A3 = TWO of u. A4 = THREE of u. M5 = ONE of r. M6 = TWO of r. E7 = ONE of t, with minor errors ignored E8 = 2 of t (a) (i) (a) (ii) (b) (i) (b) (ii) da = t 3 dt when t=4, rate = 73.2 km 2 /day da = t 3 = 0 dt t=-5 or 5, after 5 days, A=562.5km 2 dy dx = x2 7x decreasing between 1 and 6 decreasing the fastest where dy dx = x2 7x + 6 reaches a min y'' = 2x 7 = 0 x = 3.5 differentiates and finds the rate equates the derivative to 0 and finds t=5 derivative and an equation or inequation formed derives to get y derives and finds the maximum Area gives interval for x where the function is decreasing gives y and equates to zero derives and finds the x value where the gradient function is minimum (c) 100 = 2x + 2y so y = 50 x V = πr 2 x, since y = 2πr then r = y 2π forms the equation for V, derives forms the equation, derives, and solves the problem, both dimensions given.
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